Abstract Nonsense

Crushing one theorem at a time

Munkres Chapter 2 Section 19 (Part II)


9.

Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family \left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}} of non-empty sets, with \mathcal{A}\ne\varnothing, the product \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing

Proof: This is pretty immediate when one writes down the actual definition of the product, namely:

\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=\left\{\bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}:\bold{x}(\alpha)\in U_\alpha,\text{ for every }\alpha\in\mathcal{A}\right\}

So, if one assumes the AOC then one must assume the existence of a choice function

\displaystyle c:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }c\left(U_\alpha\right)\in U_\alpha\text{ for all }\alpha\in\mathcal{A}

So, then if we consider \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}\Omega as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$

i:\mathcal{A}\to\Omega

where clearly since we have already indexed out set we have that i:\alpha\mapsto U_\alpha. So, consider

c\circ i:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_\alpha

This is clearly a well-defined mapping and \left(c\circ i\right)(\alpha)=c\left(U_\alpha\right)\in U_\alpha and thus

\displaystyle c\circ i\in\prod_{\alpha\in\mathcal{A}}U_\alpha

from where it follows that \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing

Conversely, let \Omega be a class of sets and let i:\mathcal{A}\to\Omega be an indexing function. We may then index \Omega by \Omega=\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}. Then, by assumption

\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing

Thus there exists some

\displaystyle \bold{x}\in\prod_{\alpha\in\mathcal{A}}U_\alpha

Such that

\displaystyle \bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }\bold{x}(\alpha)\in U_\alpha

Thus, we have that

\displaystyle \bold{x}\circ i^{-1}:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}}

is a well-defined mapping with

\displaystyle \bold{x}\left(U_\alpha\right)\in U_\alpha

For each \alpha\in\mathcal{A}. It follows that we have produced a choice function for \Omega and the conclusion follows. \blacksquare

Remark: We have assumed the existence of a bijective indexing function i:\mathcal{A}\to\Omega, but this is either A) a matter for descriptive set theory or B) obvious since \text{id}:\Omega\to\Omega satisfies the conditions. This depends on your level of rigor.

10.

Problem: Let A be a set; let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} be an indexed family of spaces; and let \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} be an indexed family of functions f_\alpha:A\to X_\alpha

a) Prove there is a unique coarsest topology \mathfrak{J} on A relative to whish each of the functions f_\alpha is continuous.

b) Let

\mathcal{S}_\beta=\left\{f_\beta^{-1}\left(U_\beta\right):U_\beta\text{ is ope}\text{n in}X_\beta\right\}

and let \displaystyle \mathcal{S}=\bigcup_{\alpha\in\mathcal{A}}. Prove that \mathcal{S} is a subbasis for \mathfrak{J}.

c) Show that the map g:Y\to A is continuous relative to \mathfrak{J} if and only if each map f_\alpha\circ g:Y\to X_\alpha is continuous.

d) Let \displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha be defined by the equation

f(x)=\left(f_\alpha(x)\right)_{\alpha\in\mathcal{A}}

Let Z denote the subspace of f\left(A\right) of the product space \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha. Prove taht the image under f of each element of \mathfrak{J} is an open set in Z.

Proof:

a) We first prove a lemma

Lemma: Let \mathfrak{J} be a topology on A, then all the mappings f_\alpha:A\to X_\alpha are continuous if and only if \mathcal{S}\subseteq\mathfrak{J} where \mathcal{S} is defined in part b).

Proof:Suppose that all the mappings f_\alpha:A\to X_\alpha are continuous. Then, given any open set U_\alpha\in X_\alpha we have that f_\alpha is continuous and so f_\alpha^{-1}\left(U_\alpha\right) is open and thus f_{\alpha}^{-1}\left(U_\alpha\right)\in\mathfrak{J} from where it follows that \mathcal{S}\subseteq\mathfrak{J}.

Conversely, suppose that \mathcal{S}\subseteq\mathfrak{J}. It suffices to prove that f_\alpha:A\to X_\alpha  for a fixed but arbitrary \alpha\in\mathcal{A}. So, to do this let U be open in X_\alpha then f_{\alpha}^{-1}\left(U\right)\in\mathfrak{J} and thus by assumption f_\alpha^{-1}\left(U\right)\in\mathfrak{J}; but this precisely says that f_\alpha^{-1}\left(U\right) is open in A. By prior comment the conclusion follows. \blacksquare

So, let

\mathcal{C}=\left\{\mathfrak{T}:\mathfrak{J}\text{ is a topology on }A\text{ and }\mathcal{S}\subseteq\mathfrak{J}\right\}

and let

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in\mathcal{C}}\mathfrak{T}

By previous problem \mathfrak{J} is in fact a topology on A, and by our lemma we also know that all the mappings f_\alpha:A\to X_\alpha are continuous since \mathcal{S}\subseteq\mathfrak{J}. To see that it’s the coarsest such topology let \mathfrak{U} be a topology for which all of the f_\alpha:A\to X_\alpha are continuous. Then, by the other part of our lemma we know that \mathcal{S}\subseteq\mathfrak{U} and thus \mathfrak{U}\in\mathcal{C}. So,

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}\subseteq\mathfrak{U}

And thus \mathfrak{J} is coarser than \mathfrak{U}.

The uniqueness is immediate.

b) It follows from the previous problem that we must merely show that \mathcal{S} is a subbasis for the topology \mathfrak{J}. The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons):

Lemma: Let X be a set and \Omega be a subbasis for a topology on X. Then, the topology generated by \Omega equals the intersection of all topologies which contain \Omega.

Proof: Let

\mathcal{C}\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\Omega\subseteq\mathfrak{T}\right\}

and

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}

Also, let \mathfrak{G} be the topology generated by the subbasis \Omega.

Clearly since \Omega\subseteq\mathfrak{G} we have that \mathfrak{J}\subseteq\mathfrak{G}.

Conversely, let U\in\mathfrak{G}. Then, by definition to show that U\in\mathfrak{J} it suffices to show that U\in\mathfrak{T} for a fixed but arbitrary \mathfrak{T}\in\mathcal{C}. To do this we first note that by definition that

\displaystyle U=\bigcup_{\alpha\in\mathcal{A}}U_\alpha

where each

U_\alpha=O_1\cap\cdots\cap O_{m_\alpha}

for some O_1,\cdots,O_{m_\alpha}\in\Omega. Now, if \mathfrak{T}\in\mathcal{C} we know (since \Omega\subseteq\mathfrak{T}) that O_1,\cdots,O_{m_\alpha}\in\mathfrak{T} and thus

O_1\cap\cdots\cap O_{m_\alpha}=U_\alpha\in\mathfrak{T}

for each \alpha\in\mathcal{A}. It follows that U is the union of sets in \mathfrak{T} and thus U\in\mathfrak{T}. It follows from previous comment that \mathfrak{G}\subseteq\mathfrak{J}.

The conclusion follows. \blacksquare

The actual problem follows immediately from this.

c) So, let g:Y\to A be some mapping and suppose that f_\alpha\circ g:Y\to X_\alpha is continuous for each \alpha\in\mathcal{A}. Then, given a subbasic open set U in A we have that

U=f_{\alpha_1}^{-1}\left(U_{\alpha_1}\right)\cap\cdots\cap f_{\alpha_n}^{-1}\left(U_{\alpha_n}\right)

for some \alpha_1,\cdots,\alpha_n and for some open sets U_{\alpha_1},\cdots,U_{\alpha_n} in X_{\alpha_1},\cdots,X_{\alpha_n} respectively. Thus g^{-1}(U) may be written as

\displaystyle g^{-1}\left(\bigcup_{j=1}^{n}f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}g^{-1}\left(f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}\left(f_{\alpha_j}\circ g\right)^{-1}\left(U_{\alpha_j}\right)

but since each f_{\alpha_j}\circ g:Y\to X_{\alpha_j} we see that g^{-1}\left(U\right) is the finite union of open sets in Y and thus open in Y. It follows that g is continuous.

Conversely, suppose that g is continuous then f_\alpha\circ g:Y\to X_{\alpha} is continuous since it’s the composition of continuous maps.

d) First note that

\displaystyle f^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\bigcap_{\alpha\in\mathcal{A}}f\alpha^{-1}\left(U_\alpha\right)

from where it follows that the initial topology under the class of maps \{f_\alpha\} on A is the same as the initial topology given by the single map f. So, in general we note that if X is given the initial topology determined by f:X\to Y then given an open set f^{-1}(U) in X we have that f\left(f^{-1}(U)\right)=U\cap f(X) which is open in the subspace f(X).

June 9, 2010 Posted by | Fun Problems, Munkres, Topology | , , , , , , , | 1 Comment

Munkres Chapter 2 Section 19 (Part I)


1.

Problem: Suppose that for each \alpha\in\mathcal{A} the topology on X_\alpha is given by a basis \mathfrak{B}_\alpha. The collection of all sets of the form

\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha

Such that B_\alpha\in\mathfrak{B}_\alpha is a basis for \displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha with the box topology, denote this collection by \Omega_B. Also, the collection \Omega_P of all sets of the form

\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha

Where B_\alpha\in\mathfrak{B}_\alpha for finitely many \alpha and B_\alpha=X_\alpha otherwise is a basis for the product topology on X.

Proof: To prove the first part we let U\subseteq X be open. Then, by construction of the box topology for each (x_\alpha)_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}(x_\alpha)\in U we may find some \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha such that U_\alpha is open in X_\alpha and \displaystyle (x_\alpha)\in \prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U. So, then for each x_\alpha we may find some B_\alpha\in\mathfrak{B}_\alpha such that x_\alpha\in B_\alpha\subseteq U_\alpha and thus

\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}B_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U

Noticing that \displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha\in\Omega_B and every element of \Omega_B is open finishes the argument.

Next, we let U\subseteq X be open with respect to the product topology. Once again for each (x_\alpha)\in U we may find some \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha such that U_\alpha is open in X_\alpha for each \alpha\in\mathcal{A} and U_\alpha=X_\alpha for all but finitely many \alpha, call them \alpha_1,\cdots,\alpha_m. So, for each \alpha_k,\text{ }k=1,\cdots,m we may find some B_k\in\mathfrak{B}_k such that x\in B_k\subseteq U_k and so

\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}V_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U

Where

\displaystyle V_\alpha=\begin{cases}B_k\quad\text{if}\quad \alpha=\alpha_k ,\text{ }k=1,\cdots,m\\ X_\alpha\quad\text{if}\quad x\ne \alpha_1,\cdots,\alpha_m\end{cases}

Noting that \displaystyle \prod_{\alpha\in\mathcal{A}}V_\alpha\in\Omega_P and \Omega_P is a collection of open subsets of X finishes the argument. \blacksquare

2.

Problem: Let \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of topological spaces such that U_\alpha is a subspace of X_\alpha for each \alpha\in\mathcal{A}. Then, \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=Y is a subspace of \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X if both are given the product or box topology.

Proof: Let \mathfrak{J}_S,\mathfrak{J}_P denote the topologies Y inherits as a subspace of X and as a product space respectively. Note that \mathfrak{J}_S,\mathfrak{J}_P are generated by the bases \mathfrak{B}_S=\left\{B\cap Y:B\in\mathfrak{B}\right\} (where \mathfrak{B} is the basis on X with the product topology), and

\displaystyle \mathfrak{B}_P=\left\{\prod_{\alpha\in\mathcal{A}}O_\alpha:O_\alpha\text{ is open in }U_\alpha,\text{ and equals }U_\alpha\text{ for all but finitely many }\alpha\right\}

So, let (x_\alpha)\in B where B\in\mathfrak{B}_S then

\displaystyle B=Y\cap\prod_{\alpha\in\mathcal{A}}V_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)

Where V_\alpha is open in X_\alpha, and thus V_\alpha\cap Y is open in U_\alpha. Also, since V_\alpha=X_\alpha for all but finitely many \alpha it follows that V_\alpha\cap U_\alpha=U_\alpha for all but finitely many \alpha. And so B\in\mathfrak{B}_P. Similarly, if B\in\mathfrak{B}_P then

\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha

Where O_\alpha is open in U_\alpha, but this means that O_\alpha=V_\alpha\cap U_\alpha for some open set V_\alpha in X_\alpha and so

\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}V_\alpha\cap Y\in\mathfrak{B}_S

From where it follows that \mathfrak{B}_S,\mathfrak{B}_P and thus \mathfrak{J}_S,\mathfrak{J}_P are equal.

The case for the box topology is completely analgous. \blacksquare

3.

Problem: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of Hausdorff spaces, then \displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha is Hausdorff with either the box or product topologies

Proof: It suffices to prove this for the product topology since the box topology is finer.

So, let (x_\alpha),(y_\alpha)\in X be distinct. Then, x_\beta\ne y_\beta for some \beta\in\mathcal{A}. Now, since X_\beta is Hausdorff there exists disjoint open neighborhoods U,V of x_\beta,y_\beta respectively. So, \pi_\beta^{-1}(U),\pi_\beta^{-1}(V) are disjoint open neighborhoods of (x_\alpha),(y_\alpha) respectively. The conclusion follows. \blacksquare

4.

Problem: Prove that \left(X_1\times\cdots\times X_{n-1}\right)\times X_n\overset{\text{def.}}{=}X\approx X_1\times\cdots\times X_n\overset{\text{def.}}{=}Y.

Proof: Define

\displaystyle \varphi:X\to Y:((x_1,\cdots,x_{n-1}),x_n)\mapsto (x_1,\cdots,x_n)

Clearly this is continuous since \pi_{\beta}\circ\varphi=\pi_\beta

5.

Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it?

Proof: If \displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha where the latter is given the box topology then we have that each \pi_\alpha is continuous and thus so is each \pi_\alpha\circ f:A\to X_\alpha. \blacksquare#

6.

Problem: Let \left\{\bold{x}_n\right\}_{n\in\mathbb{N}} be a sequence of points in the product space \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X. Prove that \left\{\bold{x}_n\right\}_{n\in\mathbb{N}} converges to \bold{x} if and only if the sequences \left\{\pi_\alpha(\bold{x}_n)\right\}_{n\in\mathbb{N}} coverge to \pi_\alpha(\bold{x}) for each \alpha\in\mathcal{A}. Is this fact true if one uses the box topology?

Proof: Suppose that U is a neighborhood of \pi_{\alpha}(\bold{x}) such that

\left(X_\alpha-U\right)\cap\left\{\bold{x}_n:n\in\mathbb{N}\right\}=K

is infinite. Notice then that if \pi_{\alpha}(\bold{x}_n)\in K that \bold{x}_n\notin\pi_{\alpha}^{-1}\left(U\right) from where it follows that \pi_{\alpha}^{-1}\left(U\right) is a neighborhood of \bold{x} which does not contain all but finitely many values of \left\{\bold{x}_n:n\in\mathbb{N}\right\} contradicting the fact that \bold{x}_n\to\bold{x} in X.

Conversely, suppose that \pi_{\alpha}(\bold{x}_n)\to\pi_{\alpha}(\bold{x}) for each \alpha\in\mathcal{A}  and let \displaystyle \prod_{\alpha\in\mathcal{A}}U_{\alpha} be a basic open neighborhood of \bold{x}. Then, letting \alpha_1,\cdots,\alpha_m be the finitely many indices such that U_{\alpha_k}\ne X_k. Since each \pi_{\alpha_k}(\bold{x}_n)\to\pi_{\alpha_k}(\bold{x}) there exists some n_\ell\in\mathbb{N} such that n_\ell\leqslant n\implies \pi_{\alpha_k}(\bold{x}_k)\in U_k. So, let N=\max\{n_1,\cdots,n_k\}. Now, note that if \displaystyle \bold{x}_n\notin\prod_{\alpha\in\mathcal{A}}U_\alpha then \pi_{\alpha}(\bold{x}_n)\notin U_\alpha for some \alpha\in\mathcal{A}. But, since clearly \pi_{\alpha}(\bold{x}_n)\in X_\alpha we must have that \pi_{\alpha}(\bold{x}_n)\notin U_{\alpha_1}\cup\cdots\cup U_{\alpha_m} and thus n\leqslant N. It follows that for every N\leqslant n we have that \displaystyle \bold{x}_n\in\prod_{\alpha\in\mathcal{A}}U_\alpha. Then, since every neighborhood of \bold{x} contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows.

Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider \displaystyle \bold{x}_n=\prod_{m\in\mathbb{N}}\left\{\frac{1}{n}\right\}. Clearly each coordinate converges to zero, but \displaystyle \prod_{m\in\mathbb{N}}\left(\frac{-1}{m},\frac{1}{m}\right)=U is a neighborhood of \bold{0} in the product topology. But, if one claimed that for every n\geqslant N (for some N\in\mathbb{N} that \bold{x}_n\in U they’d be wrong. To see this merely note \displaystyle \frac{1}{N}\notin\left(\frac{-1}{N+1},\frac{1}{N+1}\right) and so \pi_{N+1}\left(\bold{x}_N\right)\notin\pi_{N+1}(U) and thus \bold{x}_{N}\notin U.

7.

Problem: Let \mathbb{R}^{\infty} be the subset of \mathbb{R}^{\omega} consisting of all eventually zero sequences. What is \overline{\mathbb{R}^{\infty}} in the box and product topology?

Proof: We claim that in the product topology \overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}. To see this let \displaystyle \prod_{n\in\mathbb{N}}U_n be a basic non-empty open set in \mathbb{R}^{\omega} with the product topology. Since we are working with the product topology we know there are finitely many indices n_1,\cdots,n_m such that U_{n_k}\ne \mathbb{R}. So, for each n_1,\cdots,n_m select some x_{n_k}\in U_{n_k} and consider (x_n)_{n\in\mathbb{N}} where

\displaystyle \begin{cases} x_{n_k}\quad\text{if}\quad n=n_1,\cdots,n_m \\ 0 \text{ } \quad\text{if}\quad \text{otherwise}\end{cases}

Clearly then \displaystyle (x_n)_{n\in\mathbb{N}}\in \prod_{n\in\mathbb{N}}U_n\cap\mathbb{R}^{\infty} and thus every non-empty open set in \mathbb{R}^{\omega} intersects \mathbb{R}^{\infty} and the conclusion follows.

Now, we claim that with the box topology that \overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}. To see this let (x_n)_{n\in\mathbb{N}}\notin\mathbb{R}^{\infty}. Then, there exists some subsequence \{x_{\varphi(n)}\} of the sequence \{x_n\} which is non-zero. For each \varphi(n) form an interval I_{\varphi(n)} such that 0\notin I_{\varphi(n)}. Then, consider

\displaystyle \prod_{n\in\mathbb{N}}U_n,\text{ }U_n=\begin{cases} I_{\varphi(m)}\quad\text{if}\quad n=\varphi(m)\\ \mathbb{R}\quad\text{ }\quad\text{if}\quad \text{otherwise}\end{cases}

Clearly then \displaystyle \prod_{n\in\mathbb{N}}U_n is a neighborhood of (x_n)_{n\in\mathbb{N}} and since each clearly has an infinite subsequence of non-zero values it is disjoint from \mathbb{R}^{\infty}. It follows that in \mathbb{R}^{\omega} with the box topology that \mathbb{R}^{\infty} is closed and thus \overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty} as desired. \blacksquare

8.

Problem: Given sequences (a_n)_{n\in\mathbb{N}} and (b_n)_{n\in\mathbb{N}} of real numbers with a_n>0 define \varphi:\mathbb{R}^{\omega}\to\mathbb{R}^{\omega} by the equation

\varphi:(x_n)_{n\in\mathbb{N}}\mapsto \left(a_n x_n+b_n\right)_{n\in\mathbb{N}}

Show that if \mathbb{R}^{\omega} is given the product topology that \varphi is a homeomorphism. What happens if \mathbb{R}^{\omega} is given the box topology?

Proof: Let us first prove that \varphi is a bijection. To do this we prove something more general…

Lemma: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} and \left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}} be two classes of untopologized sets and \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} a collection of bijections f_\alpha:X_\alpha\to Y_\alpha. Then, if

\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}

we have that \varphi is a bijection.

Proof: To prove injectivity we note that if

\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\varphi\left((y_\alpha)_{\alpha\in\mathcal{A}}\right)

Then,

\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=\left(f_\alpha(y_\alpha)\right)_{\alpha\in\mathcal{A}}

And by definition of an \alpha-tuple this implies that

f_\alpha(x_\alpha)=f_\alpha(y_\alpha)

for each \alpha\in\mathcal{A}. But, since each f_\alpha:X_\alpha\to Y_\alpha is injective it follows that

x_\alpha=y_\alpha

For each \alpha\in\mathcal{A}. Thus,

(x_\alpha)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}

as desired.

To prove surjectivity we let \displaystyle (y_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}Y_\alpha be arbitrary. We then note that for each fixed \alpha\in\mathcal{A} we have there is some x_\alpha\in X_\alpha such that f_\alpha(x_\alpha)=y_\alpha. So, if \displaystyle (x_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}X_\alpha is the corresponding \alpha-tuple of these values we have that

\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}

from where surjectivity follows. Combining these two shows that \varphi is indeed a bijection. \blacksquare

Lemma: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} and \left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}} be two classes of non-empty topological spaces and \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} a corresponding class of continuous functions such that f_\alpha:X_\alpha\to Y_\alpha. Then, if \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha and \displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha are given the product topologies the mapping

\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}

is continuous.

Proof: Since the codomain is a product space it suffices to show that

\displaystyle \pi_\beta\circ\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to Y_{\beta}

is continuous for each \beta\in\mathcal{A}. We claim though that the diagram

commutes where \pi^Y_\beta and \pi^X_\beta denote the canonical projections from \displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha and \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha to Y_\beta and X_\beta respectively. To see this we merely note that

\displaystyle \pi^Y_\beta\left(\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=\pi^Y_\beta\left(\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}\right)=f_{\beta}(x_\beta)

and

f_\beta\left(\pi^X_\beta\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=f_\beta\left(x_\beta\right)

which confirms the commutativity of the diagram. But, the conclusion follows since f_\beta\circ\pi_\beta is the composition of two continuous maps (the projection being continuous since \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha is a product space).

The lemma follows by previous comment. \blacksquare

We come to our last lemma before the actual conclusion of the problem.

Lemma: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} and \left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}} be two classes of non-empty topological spaces and \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} a set of homeomorphisms with f_\alpha:X_\alpha\to Y_\alpha. Then,

\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}

is a homeomorphism if \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha and \displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha are given the product topology.

Proof: Our last two lemmas show that \varphi is bijective and continuous. To prove that it’s inverse is continuous we note that

\displaystyle \left(\left(\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}\right)\circ\varphi\right)\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\prod_{\alpha\in\mathcal{A}}\left\{f_\alpha^{-1}\left(f_\alpha(x_\alpha)\right)\right\}=(x_\alpha)_{\alpha\in\mathcal{A}}

And similarly for the other side. Thus,

\displaystyle \varphi^{-1}=\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}

Which is continuous since each f_{\alpha}^{-1}:Y_\alpha\to X_\alpha is continuous and appealing to our last lemma again. Thus, \varphi is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows. \blacksquare

Thus, getting back to the actual problem we note that if we denote T_n:\mathbb{R}\to\mathbb{R}:x\mapsto a_nx+b_n that each T_n is a homeomorphism. Thus, since it is easy to see that

\displaystyle \varphi=\prod_{n\in\mathbb{N}}T_n

we may conclude by our last lemma (since we are assuming that we are giving \mathbb{R}^{\omega} in both the domain and codomain the product topology) that \varphi is a homeomorphism.

This is also continuous if we give \mathbb{R}^{\omega} the box topology. To see this we merely need to note that \displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}f_\alpha^{-1}\left(U_\alpha\right) and thus if all of the U_\alpha are open then so are (since each f_\alpha is continuous) is each f_\alpha^{-1}(U_\alpha) and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function. \blacksquare

June 7, 2010 Posted by | Fun Problems, Munkres, Topology, Uncategorized | , , , , , , | 5 Comments

Topological groups (Direct Product and Product Spaces)


Up until now we’ve discussed how the subspace topology reacts with subgroups and how the quotient topology interacts with the quotient group, it seems like a natural progression to then discuss how the product topology reacts with the direct product. So, as in the past we firstly need to verify that the two do agree, namely:

Theorem: Let G,G' be topological groups. Then, G\times G' is a topological group with the direct product group structure and product topology.

Proof: Clearly G_1\times\cdots\times G_n is a set with both a group theoretic and topological structure and so it remains to show that the map

\alpha:(G\times \cdots\times G_n)\times (G\times \cdots\times G_n)\to G\times \cdots\times G_n

given by

\alpha:\left((g_1,\cdots,g_n),(g'_1,\cdots,g'_n)\right)\mapsto (g_1{g'_1}^{-1},\cdots,g_n{g'_n}^{-1})

is continuous. But, note that

(G\times \cdots\times G_n)\times (G\times \cdots\times G_n)\approx (G\times G)\times\cdots\times (G_n\times G_n)

And that

\alpha_G\times\cdots\times \alpha_{G_n}:(G\times G)\times\cdots\times (G_n\times G_n)

given by

\alpha:\left((g_1,g'_1),\cdots,(g_n,g'_n)\right)\mapsto\left(g_1{g'_1}^{-1},\cdots,g_n{g'_n}^{-1}\right)

being the product of continuous maps is continuous. The conclusion immediately follows. \blacksquare

There are canonical topological epimorphisms from G_1\times\cdots\times G_n into G_k. Namely:

Theorem: Let

\pi_k:G_1\times G_n\to G_k:(g_1,\cdots,g_k,\cdots,g_n)\mapsto g_k

Then, \pi_k is an open topological epimorphism.

Proof: The fact that it’s open, surjective, and continuous follows since topologically \pi_k is merely the canonical projection of a product space onto it’s kth coordinate. Thus, it remains to show that it’s a homomorphism. But, this is a routine calculation. \blacksquare

Theorem: Let \pi_k be as above. Then,

\ker\pi_k=G_1\times\cdots\times\underbrace{\{e\}}_{k}\times\cdots\times G_n

from where it follows from a previous theorem that

\left(G_1\times\cdots\times G_n\right)/\left(G_1\times\cdots\times\{e\}\times\cdots\times\right)\overset{\text{T.G.}}{\cong}G_k. \blacksquare

Now, so we can use it we should mention in passing that:

Theorem: G_1\times\cdots\times G_n\overset{\text{T.G.}}{\cong}G_{\sigma(1)}\times\cdots\times G_{\sigma(n)} where \sigma:\{1,\cdots,n\}\to\{1,\cdots,n\} is any bijeciton.

Proof: Obvious. \blacksquare

In fact, that is all I wanted to mention about them. Unfortunately the interesting stuff about the direct product of topological groups that is interesting isn’t “accessible” and would take too much time to really discuss. Thus, for once I have a post which is less than one-thousand words. Enjoy.

April 29, 2010 Posted by | Algebra, Group Theory, Topological Groups, Topology, Uncategorized | , , , | Leave a comment

The dreaded product topology (Pt. III Tychonoff’s theorem and odds and ends)


In this post we will discuss one of the most important in point-set topology, Tychonoff’s theorem, and prove up some miscellaneous facts that are either interesting or useful.

Theorem (Tychonoff): Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty class of non-empty compact topological spaces, then \displaystyle X=\prod_{j\in\mathcal{J}}X_j is compact under the product topology.

Proof: It follows from an earlier theorem that we must merely show that every class of closed sets in the defining closed subbase with the F.I.P have non-empty intersection. So, let \mathcal{S}=\left\{S_k\right\}_{k\in\mathcal{K}} be such a class. Consider some arbitrary j\in\mathcal{j}, since S_k is subbasic closed for ever S_k\in\mathcal{S} we know that \pi_j(S_k) is closed, and so \left\{\pi_j(S_k)\right\}_{j\in\mathcal{J}} is a class of closed subsets of X_j, and noticing that \pi_{j}(S_{k_1})\cap\cdots\cap\pi_j(S_{k_n})\supseteq\pi_j\left(S_{k_1}\cap\cdots\cap S_{k_n}\right)\supseteq\varnothing and so by assumption we have that \displaystyle \bigcap_{k\in\mathcal{K}}\pi_j(S_k)\ne \varnothing. In other words, we have there exists some \displaystyle x_j\in\bigcap_{k\in\mathcal{K}}\pi_j(S_k) and so every element of \mathcal{S} contains an element which has x as it’s jth coordinate. Doing this for each j\in\mathcal{J} produces a set \left\{x_j\right\}_{j\in\mathcal{J}} such that \displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\bigcap_{k\in\mathcal{K}}S_k. In light of our initial comments the conclusion follows. \blacksquare

From this we can prove a nice little result that is used incessantly in multivariate analysis. But, before we continue we need a little terminology

n-cell: An n-cell is a multi-dimensional generalization of   a closed interval in \mathbb{R}^n . In other words, we call E an n-cell if \pi_k(E)=[a,b] for every 1\leqslant k\leqslant n. In other words, E is an n-cell if there exists some [a,b]\subseteq\mathbb{R} such that E=\underbrace{[a,b]\times\cdots\times[a,b]}_{n\text{ times}}=[a,b]^n.

Theorem (Generalized Heine-Borel): Every closed and bounded subspace of \mathbb{R}^n is compact.

Proof: Let E be such a set. Since E is bounded we know that E is contained in some n-cell [a,b]^n. So, if we prove that every n-cell is compact we are done (since a closed subspace of a compact space is compact).

But, we know that [a,b]^n=[a,b]\times\cdots\times[a,b] and since [a,b]\subseteq\mathbb{R} is compact we have by Tychonoff’s theorem that the product of compact spaces is compact. The conclusion follows. \blacksquare

Be careful about taking this and attempting to try to make a generalization of the extreme value theorem, namely “$If f:X\mapsto\mathbb{R}^n is continuous and X compact then \sup\text{ }f\left(X\right),\inf\text{ }f\left(X\right)\in f\left(X\right)“. It fails for an easily over-looked and decidedly non-topological property of \mathbb{R}^n,text{ }n\geqslant 2. Particularly, we have that \mathbb{R} has a canonical ordering (the one we know and love), but what is the ordering on \mathbb{R}^n? No ordering no concept of infimums and supremums.

We now take a look at some of the odds and ends that were previously mentioned. It should probably be mentioned for those that are reading this and already know topology that, for now, we omit mention of the majority (save Hausdorfness AKA T_2) of the separability axioms and things like local compacntess, connectedness, paracompactness, and path-connectedness. We shall mention them in due time, from where we will discuss which are invariant under product.

Theorem: Let X_1,\cdots,X_n be a finite number of discrete spaces, then the product set X=X_1\times\cdots\times X_n is the same topological space under both the product and discrete topology.

Proof: Let \mathcal{J} be the product topology on X and, of course, let \wp\left(X\right) be the discrete topology (I use the \wp for power set). It is clear that \mathcal{J}\subseteq\wp\left(X\right) and so we must merely show the reverse inclusion to finish.

So, let E\in\wp\left(X\right) then E=U_1\times\cdots\times U_n for some subsets U_k\subseteq X_k, but since X_k is discrete we know that each U_k is open. Thus, E is the finite product of open sets and thus open in the product (box) topology. The conclusion follows. \blacksquare

The fact that I noted that product topology and box topology were the same here was meaningful. We know that the two coincide  for the product of finitely many spaces, but the next remark shows (once again) how they differ for infinite.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be an arbitrary collection of discrete spaces, then \displaystyle X=\prod_{j\in\mathcal{J}}X_j under the box topology coincides with X under the discrete topology. The same cannot be said about the product topology.

Proof: Follows the exact same reasoning as in the last theorem.

To see why the infinite product of discrete spaces need not be discrete under the product topology, one need only consider \displaystyle X=\prod_{j=1}^{\infty}X_j where X_j=[0,1] under the discrete topology. Clearly then (0,1)\times\cdots\times(0,1)\times\cdots is open in X under the discrete topology but NOT the product topology. \blacksquare

The last thing we will talk about is the product of maps.

Product of maps: Let f:X\mapsto Y and f':X'\mapsto Y' be two maps, then define the product of the maps to be f\times f':X\times X'\mapsto Y\times Y' by (x,x')\mapsto (f(x),f'(x')).

We extend this definition to an arbitrary number of topological spaces as follows: Let \left\{X_j\right\}_{j\in\mathcal{J}} and \left\{Y_j\right\}_{j\in\mathcal{J}} be two sets of topological spaces and let \left\{f_j\right\}_{j\in\mathcal{J}} be a class of mappings such that f_j:X_j\mapsto Y_j (this can obviously be formulated in the guise of ordered triples as well), we then define \displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{j}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j  by \displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\mapsto\prod_{j\in\mathcal{j}}\{f_j(x_j)\}

The reader may recall that this concept was used to prove that if f:X\mapsto Y is continuous and Y Hausdorff then \Gamma_f is a closed subset of X\times Y.

We begin by proving the product of continuous maps is continuous. This is the first of a sequence of three proofs which will lead to a very satisfying corollary.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} and \left\{Y_j\right\}_{j\in\mathcal{J}} be an arbitrary collection of topological spaces and let \left\{f_j\right\}_{j\in\mathcal{J}} be a corresponding set of continuous mappings such that f_j:X_j\mapsto Y_j. Then, \displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j is continuous, where \displaystyle \prod_{j\in\mathcal{J}}X_j,\prod_{j\in\mathcal{J}}Y_j are under the product topology.

Proof:

Lemma: \displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j^{-1}\left(O_j\right)

Proof: Let \displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\left( \prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right) then, \displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}\in\prod_{j\in\mathcal{J}}O_j which clearly means

that f_j(x_j)\in O_j,\text{ }\forall j\in\mathcal{J} and so x_j \in f^{-1}(O_j) and thus

\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)

Conversely, let \displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right). Then, x_j\in f^{-1}\left(O_j\right)\implies f(x_j)\in O_j,\text{ }\forall j\in\mathcal{J} and thus \displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)\in \prod_{j\in\mathcal{J}}O_j and thus \displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right). The conclusion follows. \blacksquare

Now, as was previously proven we need only prove that the inverse image of each open basic set is open. So, let B\in\mathfrak{B} (the defining open base for \displaystyle \prod_{j\in\mathcal{J}}Y_j). Then, \displaystyle B=\prod_{j\in\mathcal{J}}O_j where O_j=Y_jfor all but finitely many j, and thus \displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(B\right)=\left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right). But, we see that since O_j is an open set which equals Y_j for all but finitely many j‘s that f^{-1}(O_j) is going to be an open set (since each f_j:X_j\mapsto Y_j is continuous) which equals X_j for all but finitely many j‘s. The conclusion follows. \blacksquare

Theorem: Let everything be as in the previous theorem, except instead of each f_j:X_j\mapsto Y_j being continuous, let it be bijective. Then \displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j is bijective.

Proof:

Injectivity: Suppose that \displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{y_j\}\right). Then, by definition \displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{f_j(y_j)\} and since two functions (remember we think of arbitrary tuples as functions) are equal if and only if they take the same value at every point we see that f_j(x_j)=f_j(y_j),\text{ }\forall j\in\mathcal{J} and since each f_j is bijective (and thus injective) it follows that x_j=y_j,\text{ }\forall j\in\mathcal{J}, from where injectivity follows.

Surjectivity: Let \displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}Y_j be arbitrary. For each j\in\mathcal{J} we have that there exists some x_j\in X_j such that f_j(x_j)=y_j and thus \displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\} from where surjectivity immediately follows. \blacksquare

Theorem: Let everything be as before, except now instead of continuous or bijective assume that each f_j:X_j\mapsto Y_j is surjective and open. Then, \displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j is open.

Proof:

Lemma: \displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j(O_j).

Proof: Let \displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right) then \displaystyle \prod_{j\in\mathcal{J}}\{y_j\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}x_j\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\} for some \displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}(O_J). Clearly then we have that x_j\in O_j and so f_j(x_j)\in O_j and so \displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j(O_j).

Conversely, let \displaystyle\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}\{f_j(O_j)\} then y_j=f_j(x_j),\text{ }\forall j\in\mathcal{J} for some x_j\in O_j and thus \displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)

The conclusion follows. \blacksquare.

Now, let E be open in \displaystyle \prod_{j\in\mathcal{J}}X_j then \displaystyle E=\prod_{j\in\mathcal{J}}O_j where O_j=X_j for all but finitely many j it follows then that f_j(O_j) is an open set which equals Y_j for all but finitely many j (since each f_j:X_j\mapsto Y_j is surjective and open). Thus \displaystyle \prod_{j\in\mathcal{J}}f_j(E)=\prod_{j\in\mathcal{J}}f_J\left(\prod_{j\in\mathcal{J}}\right)=\prod_{j\in\mathcal{J}}f_j(O_j) which by prior comment is the product of open sets in \displaystyle \prod_{j\in\mathcal{J}}Y_j which equals the full space for all but finitely many j. It follows that f(E) is open in \displaystyle \prod_{j\in\mathcal{J}}Y_j under the product topology. \blacksquare
Theorem: Combining these three we see that if X_j\approx Y_j for all j\in\mathcal{J}, then there exists some homeomorphism (a bijective continuous open map)  f_j:X_j\mapsto Y_j and so \displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j is a bijective continuous open map and thus a homeomorphism. It follows that \displaystyle \prod_{j\in\mathcal{J}}X_j\approx\prod_{j\in\mathcal{J}}Y_j where each are under the product topology.

February 21, 2010 Posted by | General Topology, Topology | , , , | 1 Comment

The dreadful product topology (Pt. II)


In the previous post we gave the definitions of the product topology and some nice little tools about it. Here, we give some theorems regarding what properties of topological spaces are invariant under the formation of the product topology. Let us start out nice and easy

Theorem: Let X_1,\cdots,X_n be a finite number of topological spaces with corresponding open bases \mathfrak{B}_1,\cdots,\mathfrak{B}_n. Then, \mathfrak{M}=\left\{B_1\times\cdots\times B_n:B_k\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant m\right\} is an open base for X_1\times\cdots\times X_n under the product topology.

Proof: Let O be open in X, and let o\in O. There exists some B\in\mathfrak{B} (the defining open base) such that o\in B\subseteq O. But, by theorem 1 in the last post we know that \pi_k(B) is an open set containing \pi_k(o) for 1\leqslant k\leqslant n and since \mathfrak{B}_k is an open base for X_k we know there exists some J_k\in\mathfrak{B}_k such that \pi_k(o)\in J_k\subseteq \pi_k(B). Clearly then o\in J_1\times J_n\subseteq B and since J_1\times J_n\in\mathfrak{M} and o,O were arbitrary the conclusion follows. \blacksquare

We know use this to prove a nice little theorem.

Theorem: Let X_1,\cdots,X_n be a finite number of second countable spaces, then X=X_1\times\cdots\times X_n under the product topology is second countable.

Proof: Let \mathfrak{M} be as above. Since \mathfrak{M} is an open base for X is remains to prove that it is countable. Let \eta:\mathfrak{M}\mapsto\mathfrak{B}_1\times\cdots\times\mathfrak{B}_n be given by B_1\times\cdots\times B_n\mapsto \left(B_1,\cdots,B_n\right). It is clear that this is a bijection. Recalling that the finite product of countable sets is countable finishes the argument. \blacksquare

It turns out that the above is a little more restrictive than is necessary as the following theorem illustrates.

Theorem: Let \left\{X_n\right\}_{n\in\mathbb{N}} be a sequence of second countable spaces, then \displaystyle X=\prod_{n=1}^{\infty}X_n is second countable under the product topology.

Proof: Let \left\{\mathfrak{B}_n\right\} be the corresponding countable open bases and define \displaystyle \mathcal{B}_m=\left\{E:E=\prod_{j=1}^{\infty}G_j\right\} where \pi_k(G_j)\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant m and \pi_k(G_j)=X_j,\text{ }m<k.

It is relatively easy to prove that \eta:\mathcal{B}_m\mapsto\mathfrak{B}_1\times\cdots\times\mathfrak{B}_m given by \displaystyle B_1\times\cdots\times B_m\times\prod_{j=m+1}^{\infty}X_{j+1}\mapsto \left(B_1,\cdots,B_m\right)  is a bijection. It follows then, since the finite product of countable sets is countable, that \mathcal{B}_m is countable.

So, now define \displaystyle \mathfrak{M}=\bigcup_{j=1}^{\infty}\mathcal{B}_j. It is clear that since \mathfrak{M} is the countable union of countable sets that it is countable. So now, it remains to prove that \mathfrak{M} is in fact an open base for X

So, let O be open in X and let o\in O. There exists some B in the defining open base such that o\in B\subseteq O. Considering the form in which basic open sets in the product topology take we know that there is a finite set of indices \left\{j_1,\cdots,j_n\right\} such that \pi_{j_k}(B)\ne X_k. So, let m=\max\{j_1,\cdots,j_n\}. Clearly we have that \pi_\ell(O) is a neighborhood of \pi_k(o) for 1\leqslant k\leqslant m and since \mathfrak{B}_k is an open base there exists some B_k such that \pi_k(o)\in B_k\subseteq\pi_k(O). So, let U=B_1\times\cdots\times B_m\times\displaystyle \prod_{j=m+1}^{\infty}X_j. Then, it is evident that o\in U\subseteq B\subseteq O. Noting that U\in\mathfrak{M} finishes the argument. \blacksquare.

Note the necessity for finiteness again.

Theorem: Let X_1,\cdots,X_n be a finite number of toplogical spaces and let \mathfrak{D}_1,\cdots,\mathfrak{D}_n be a number of corresponding dense sets. Then, \mathfrak{D}_1\times\cdots\times\mathfrak{D}_n is dense in X=X_1\times\cdots\times X_n under the product topology.

Proof: Let x\in X be arbitrary and let N be any neighborhood of x. We then have by theorem 1 in the last post that \pi_k(N) is a neighborhood of \pi_k(x) for 1\leqslant k\leqslant n and since \mathfrak{D}_k is dense in X_k there exists some \mathfrak{d}_k\in\mathfrak{D}_k such that \mathfrak{d}_k\in\pi_k(N). Clearly then \left(\mathfrak{d}_1,\cdots,\mathfrak{d}_n\right)\in N, and clearly \left(\mathfrak{d}_1,\cdots,\mathfrak{d}_n\right)\in\mathfrak{D}_1\times\cdots\times\mathfrak{D}_n. Noting the arbitrariness of x, N finishes the argument. \blacksquare

From this we can derive the obvious consequence.

Theorem: Let X_1,\cdots,X_n be finite number of separable spaces and let X=X_1\times\cdots\times X_n under the product topology, then X is separable.

Proof: Since each X_k is separable it has a countable dense subset \mathfrak{D}_k and by the above theorem we see that \mathfrak{D}_1\times\cdots\times\mathfrak{D}_n is dense in X, but since the finite product of countable sets is countable the conclusion follows. \blacksquare

As in the case of second countable spaces we may extend this to the countable case

Theorem: Let \left\{X_n\right\}_{n\in\mathbb{N}} be sequence of separable spaces, then \displaystyle X=\prod_{j=1}^{\infty}X_j is separable.

Proof: Since each X_n is separable it has a countable dense subset \mathfrak{D}_n we may extract an arbitrary but fixed sequence from them as \left\{d_n\right\}_{n\in\mathbb{N}} where d_n\in\mathfrak{D}_n. We now define \displaystyle \mathcal{D}_m=\prod_{j=1}^{\infty}G_j where G_j=\begin{cases} \mathfrak{D}_j & \mbox{if} \quad 1\leqslant j\leqslant m \\ \{d_j\} & \mbox{if} \quad j>m\end{cases}

Clearly we have that \eta:\mathcal{D}_m\mapsto\mathfrak{D}_1\times\cdots\times\mathfrak{D}_m by D\mapsto \left\{\pi_1(D)\right\}\times\cdots\times \left\{\pi_m(D)\right\} is a bijection and since the finite product of countable sets is countable it follows that \mathcal{D}_m is countable. So, let \displaystyle \mathcal{D}=\bigcup_{m=1}^{\infty}\mathcal{D}_m being the countable union of countable sets is countable.

From the previous paragraph it is clear that we must merely show that \mathcal{D} is dense in X. To do this let x\in X be arbitrary and N any neighborhood of X. There exists some B\in\mathfrak{B} (the defining open base) such that x\in B\subseteq N. But, we know that \pi_{k}(B)\ne X_k for finitely man indices, say k_1,\cdots,k_m. Let r=\max\{k_1,\cdots,k_m\}. We know then that \pi_\ell(B) is a neighborhood of \pi_\ell(x) for every 1\leqslant \ell\leqslant r and since \mathfrak{D}_\ell is dense in X_\ell there exists some \mathfrak{d}_\ell such that \mathfrak{d}_\ell\in\pi_\ell(B). Clearly then, we have that \displaystyle D=\prod_{j=1}^{\infty}\{g_j\} where g_j=\begin{cases} \mathfrak{d}_j & \mbox{if} \quad 1\leqslant j\leqslant r\\ d_j & \mbox{if} \quad j>r\end{cases} is in B and thus N. But, D\in\mathcal{D} and the conclusion follows. \blacksquare

We now discuss the product of Hausdorff spaces.

Theorem: \left\{X_j\right\}_{j\in\mathcal{J}} be an arbitrary collection of Hausdorff spaces, then \displaystyle X=\prod_{j\in\mathcal{J}}X_j under the product topology is Hausdorff.

Proof: Let \bold{x},\bold{y}\in X such that \bold{x}\ne \bold{y} then \pi_k(\bold{x})\ne\pi_k(\bold{x}) for some k\in\mathcal{J} and since X_k is Hausdorff we know that there exists open sets U,V such latex \pi_k(\bold{x})\in U,\pi_k(\bold{y})\in V and U\cap V. So define \displaystyle E=\prod_{j\in\mathcal{J}}G_j where G_j=\begin{cases} U & \mbox{if}\quad j=k \\ X_j & \mbox{if}\quad k\ne j\end{cases} and define \displaystyle F=\prod_{j\in\mathcal{J}}H_j where H_j=\begin{cases} V & \mbox{if}\quad j=k \\ X_j & \mbox{if} \quad x\ne j\end{cases} clearly we have that \bold{x}\in E and \bold{y}\in F and since \pi_k(F)\cap\pi_k(E)=\varnothing we must have F\cap E=\varnothing. The conclusion follows. \blacksquare

February 20, 2010 Posted by | General Topology, Topology | , , | 4 Comments

The dreaded product topology


NOTE: All sets and spaces in the following discussion will be assumed to be, without loss of generality, non-empty. Also, if not stated it is assumed that \displaystyle X=\prod_{j\in\mathcal{J}}X_j under the product topology where \left\{X_j\right\}_{j\in\mathcal{J}} is a non-empty collection of topological spaces.

To continue any further we mus make an, admittedly not brief, sojourn into the concept of product topologies.  Now, this isn’t a text-book so I’ll assume that the reader is familiar with the basic concepts of Cartesian products and projections (denote \pi). With this in mind we first make a definition for the arbitrary Cartesian product of sets:

Arbitrary Cartesian Product: Let \left\{X_j\right\}_{j\in\mathcal{J}} be an arbitrary class of non-empty sets, then we define \displaystyle \prod_{j\in\mathcal{J}}X_j=\left\{\bold{x}:\mathcal{J}\mapsto\bigcup_{j\in\mathcal{J}}X_j \text{ }:\text{ } \bold{x}(j)\in X_j,\text{ }\forall j\in\mathcal{J}\right\}.

Now, if \left\{X_j\right\}_{j\in\mathcal{J}} is a non-empty class of topological spaces we define the product topology on \displaystyle X=\prod_{j\in\mathcal{J}}X_j as the topology generated by the subbase \mathfrak{S}=\left\{\pi_j^{-1}(G): G \in\mathfrak{J}_j\right\} (where \mathfrak{J}_j is the topology on X_j,) in the sense that it is the set of all finite intersections of elements of \mathfrak{S} form an open base \mathfrak{B} for X, and thus the topology on X is made up of all arbitrary unions of elements of this open base. We call \mathfrak{S} in the above discussion the defining open subbase for X and \mathfrak{B} the defining open base for X under the product topology.

So what do elements of \mathfrak{S} look like? Well, let G be an open set in X_j then \pi_j^{-1}(G) is (in words) all \bold{x}\in X such that \pi_j(\bold(x))\in G. It is clear that the only limitation then that needs to be made on \bold{x} is that it’s jth coordinate is in G. From this it is clear that \displaystyle \pi_j^{-1}(G)=\prod_{\ell\in\mathcal{J}}E_\ell where \displaystyle \begin{cases} X_j & \mbox{if} \quad \ell\ne j\\ G & \mbox{if} \quad \ell=j\end{cases}. Thus, \mathfrak{S} is made up of all sets of the form \displaystyle \prod_{j\in\mathcal{J}}E_j where E_j=X_j for all but one j

Example: If X_j=[0,1],\text{ }j\in[0.1] under the usual topology then \displaystyle (0,1)\times\prod_{j\in(0,1]}X_j\in\mathfrak{S} whereas \displaystyle (0,1)\times\prod_{j\in(0,1)}X_j\times(0,1)\notin\mathfrak{S}

So, this being said we see that the elements of \mathfrak{B} (being the finite intersection of elements of \mathfrak{S}) are going to be the product of open sets G_j\subseteq X_j such that G_j=X_j for all but finitely many j.

So, let’s start proving some theorems!

Theorem 1: All projections \pi_j:X\mapsto X_j are open mappings.

Proof: Let E\subseteq X be open then \displaystyle E=\bigcup_{k\in\mathcal{K}}B_k where \left\{B_k\right\}_{k\in\mathcal{K}}\subseteq\mathfrak{B}, and so \displaystyle \pi_j(E)=\pi_j\left(\bigcup_{k\in\mathcal{K}}B_k\right)=\bigcup_{k\in\mathcal{K}}\pi_j\left(B_k\right) and since the jth coordinate of every basic open set (whether equal to the full space or not) is an open set in X_j and so \pi_j(E) is the union of open sets inX_j and thus open. \blacksquare

Corollary 1: If N is a neighborhood of \bold{x}\in X then \pi_j(N) is a neighborhood of \pi_j(\bold{x})

Remark: Projections need not be closed mappings. Consider the set E=\left\{(x,y):\text{ }xy=1,\right\} as a subset of \mathbb{R}^2 with the product topology. It is relatively easy E is closed, but \pi_1(E)=\pi_2(E)=\mathbb{R}-\{0\} which is open.

One might ask “Why not just make the open base for a topology on X the set of all arbitrary products of open sets?”. The topology that this is describing is called the box topology. Namely, the box topology has the set of all arbitrary products of open open sets as an open base. It is clear that the box and product topologies coincide for finite cases but do not for infinite cases. So, why did we choose the box topology over the product topology? The answer comes from the following theorem.

Theorem: Let f:M\mapsto X be a mapping of a topological space M into the product space X. Then, f is continuous if and only if \pi_jf:M\mapsto X_j is continuous for all j.

Proof:

\implies:

Let G be open in X_j then \left(\pi_j f\right)^{-1}\left(G\right)=f^{-1}\left(\pi_j^{-1}(G)\right) and since \pi_j^{-1}(G) is subbasic open in X and f continuous we see that f^{-1}\left(\pi_j^{-1}(G)\right) is open in M.

\Leftarrow:

Lemma: A mapping f:E\mapsto F (where E and F are arbitrary topological spaces) is continuous if and only if f^{-1}(B) is open for every B\in\mathcal{B}_F where \mathcal{B}_F is an open base for F.

Proof:

\implies: Let G be open in F then \displaystyle G=\bigcup_{h\in\mathcal{H}}B_h where \left\{B_h\right\}_{h\in\mathcal{H}}\subseteq\mathcal{B}_F. So then, \displaystyle f^{-1}\left( G\right)=\displaystyle f^{-1}\left(\bigcup_{h\in\mathcal{H}}B_h\right)=\bigcup_{h\in\mathcal{H}}f^{-1}\left(B_h\right) which by assumption is the union of open sets in E and thus open. The conclusion follows.

\Leftarrow: This is obvious. \blacksquare

So, suppose that G is basic open in X, then \displaystyle G=\pi_{j_1}^{-1}(G_{j_1})\cap\cdots\cap\pi_{j_m}^{-1}(G_{j_m}) where \left\{j_,\cdots,j_m\right\}\subseteq\mathcal{J} and G_{j_k} is open in X_k,\text{ }1\leqslant k\leqslant m. It follows then that

f^{-1}(G)

=f^{-1}\left(\pi_{j_1}^{-1}(G_{j_1})\cap\cdots\cap\pi_{j_m}^{-1}(G_{j_m})\right)

=f^{-1}\left(\pi_{j_1}^{-1}(G_{j_1})\right)\cap\cdots\cap f^{-1}\left(\pi_{j_m}^{-1}(G_{j_m})\right)

=\left(\pi_{j_1}f\right)^{-1}(G_{j_1})\cap\cdots\cap \left(\pi_{j_m}f\right)^{-1}(G_{j_m})

and since each \pi_jf,\text{ }j\in\mathcal{J} is continuous it follows that f^{-1}(G) is the finite intersection of open sets in M and thus open. The conclusion follows. \blacksquare

Note that the key point here was that G was the finite intersection of inverse images of open sets. If we had the box topology we could have had that G was the infinite intersection of inverse images of open sets we would have no guarantee that f^{-1}(G) would be open since it would be the infinite intersection of open sets…which we all know need not be open.

Now, one may ask “Topologically does it matter which order we product the sets?”. The answer is “no…for finite cases”

Theorem: Let X_1,\cdots,X_n be a finite collection of topological spaces then X_1\times\cdots\times X_n\approx X_{\sigma(1)}\times\cdots\times X_{\sigma(n)} where \sigma is an permutation of \{1,\cdots,n\}

Proof: It is readily verified that the canonical mapping f:X_1\times\cdots\times X_n\mapsto X_{\sigma(1)}\times\cdots\times X_{\sigma(n)} given by (x_1,\cdots,x_n)\mapsto (x_{\sigma(1)},\cdots,x_{\sigma(n)}) is a homeomorphism. \blacksquare

February 20, 2010 Posted by | General Topology, Topology | , | 4 Comments